UPSC 2015 CDS I Math Paper

Let the distance between the parallel sides be h cm.
ΔABE is an equilateral triangle.
∠A = ∠B = ∠E = 60°
Also, ∠EDC = ∠A=60° and ∠ECD = ∠B=60°  (Corresponding pair of angles)
Therefore,ΔEDC is an equilateral triangle
AE = BE
DE = CE
∴ AE- DE = BE - CE
⇒AD = BC ...(1)
Therefore, trapezium ABCD is an isosceles trapezium.
⇒BD = AC  (Diagonals of isosceles trapezium are equal)
Now ΔAPD ≅ ΔBQC (RHS congruence rule)
∴ AP = BQ = b(say)
Let CD = PQ = a
Area of trapezium
=

1

2

×AC×OD+

1

2

×AC×OB
=

1

2

×AC×BD
=

1

2

BD2
Using (1)
∴

1

2

BD2=16
⇒BD2=32
Area of trapezium
∴(a+b)×h=16
⇒a+b=

16

h

In right ΔBPD,
h2+(a+b)2=BD2
⇒h2+(

16

h

)2=32
Using option:
⇒h=4cm